// Numbas version: exam_results_page_options {"name": "Parametric 2nd derivative", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Parametric 2nd derivative", "tags": [], "metadata": {"description": "Calculate d2y/dx2 for a curve defined parametrically", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A curve is defined by the parametric equations $x=\\var{a}+\\var{b}t^2, y = \\var{c}t+\\var{d}t^3$

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We have $x=\\var{a}+\\var{b}t^2, y = \\var{c}t+\\var{d}t^3$

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So $\\frac{dx}{dt} = 2 \\times\\var{b}t = \\var{2b}t$

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and $\\frac{dy}{dt} = \\var{c}+3 \\times\\var{d}t^2 = \\var{c}+\\var{3d}t^2$

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Since $\\frac{dy}{dx}=\\frac{dy}{dt}/\\frac{dx}{dt}$ we have $\\frac{dy}{dx}=\\frac{\\var{c}+\\var{3d}t^2}{\\var{2b}t}=\\frac{\\var{c}}{\\var{2b}t}+\\frac{\\var{3d}t}{\\var{2b}}$

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To find $\\frac{d^2y}{dx^2}$ we use the chain rule to obtain $\\frac{d^2y}{dx^2}=\\frac{d}{dx}(\\frac{dy}{dx})=\\frac{d}{dt}(\\frac{dy}{dx})\\times\\frac{dt}{dx}=\\frac{\\frac{d}{dt}(\\frac{dy}{dx})}{\\frac{dx}{dt}}$

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Now $\\frac{d}{dt}(\\frac{dy}{dx}) = \\frac{-\\var{c}}{\\var{2b}t^2}+\\frac{\\var{3d}}{\\var{2b}}=\\frac{-\\var{c}+\\var{3d}t^2}{\\var{2b}t^2}$

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Hence $\\frac{d^2y}{dx^2}=\\frac{\\frac{d}{dt}(\\frac{dy}{dx})}{\\frac{dx}{dt}}=\\frac{-\\var{c}+\\var{3d}t^2}{\\var{2b}t^2\\times\\var{2b}t}=\\frac{\\var{3d}t^2-\\var{c}}{\\var{4b^2}t^3}$

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Find $\\frac{d^2y}{dx^2}$ in terms of $t$    [[0]]

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